Problem 666

If (x,y) is a point on the unit circle in quadrant IV and if x = $\frac{\sqrt{2}}{2}$ , what is y?

Solution:-

If the point (x,y) is on the unit circle, then the point satisfies the equation of the unit circle

$x^{2}+y^{2}=1$

$(\frac{\sqrt{2}}{2})^{2}+y^{2}$ =1

Evaluate $(\frac{\sqrt{2}}{2})^{2}$

$\frac{1}{2}+ y^{2} = 1$

Solve for y.

$y^{2} = \frac{1}{2}$

Take the square root of both sides.

y = $\pm \sqrt{\frac{1}{2}}$

y = $\pm \frac{\sqrt{2}}{2}$

In quadrant IV, the y-coordinate of the point is negative. Therefore, y = – $\frac{\sqrt{2}}{2}$